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Question

One mole of an ideal gas (monatomic) is taken from A to C along path ABC. The temperature of gas is T0 at A, then for process ABC

A
Work done by gas is RT0.
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B
Change in internal energy is 112RT0.
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C
Heat absorb by gas is 112RT0.
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D
Heat absorb by gas is 132RT0.
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Solution

The correct options are
A Work done by gas is RT0.
C Heat absorb by gas is 112RT0.
Area under PV graph= work done=P0V0=RT0

P0V0T0=2P0.2V0TCTC=4T0

Since, ΔU=nCVΔT
ΔU=1×32R(4T0T0)ΔU=92RT0Q=ΔU+WBy putting valuesQ=112RT0

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